Question

12.16.  Randomly selected groups of 120 parents and 150 teachers from one school district are surveyed about their attitudes toward inclusion. One of the questions asks them whether they oppose or support inclusions and their responses to this question are recorded in the following table. The data were analyzed using a chi square test. The obtained chi square value is 5.65, significant at the .02 level (p=.02).

GROUP	SUPPORT	OPPOSE
Parent	75	45
Teachers	72	78

a.             Which chi square test should be used to analyze the data and answer the research questions? Explain.

b.             Is there a statistically significant difference in the responses of the parents and teachers? Explain.

12.17.  In a recent national poll, people were asked the following question: "In your opinion, how important is it to improve the nation's inner-city schools?" The responses of city residents who do not have school-age children were compared to the national responses. A chi square test was used to analyze the data in order to determine whether there is a difference in responses between those who live in cities and do not have school-age children and the national responses. The results of the study are displayed in the following table. The analysis revealed a chi square value of 4.32, significant at p=.36.

RESPONSE	NO CHILDREN IN SCHOOL	NATIONAL TOTALS
Very Important	78	80
Fairly Important	13	15
Not VeryImportant	6	3
Not Important at All	2	1
Don't Know	1	1

a.         Which chi square test was used to analyze the data? Explain.

b.        What was the null hypothesis?

c.         What are the conclusions of the study? Explain.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: There is no difference in responses between those who live in cities and do not have school-age children and the national responses.

Alternative hypothesis: There is difference in responses between those who live in cities and do not have school-age children and the national responses.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for homogeneity.

Analyze sample data. Applying the chi-square test for homogeneity to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = (r - 1) * (c - 1) = (5 - 1) * (2 - 1)
D.F = 4
E_r,c = (n_r * n_c) / n

?² = 265

where DF is the degrees of freedom, r is the number of populations, c is the number of levels of the categorical variable, n_r is the number of observations from population r, n_c is the number of observations from level c of the categorical variable, n is the number of observations in the sample, E_r,c is the expected frequency count in population r for level c, and O_r,c is the observed frequency count in population r for level c.

The P-value is the probability that a chi-square statistic having 4 degrees of freedom is more extreme than 265.

We use the Chi-Square Distribution Calculator to find P(?² > 265) = less than 0.0001

Interpret results. Since the P-value (0.0000) is less than the significance level (0.05), we reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is difference in responses between those who live in cities and do not have school-age children and the national responses.